Question

Solve the following equation: $2\left( {\dfrac{{x - 1}}{{x + 3}}} \right) - 7\left( {\dfrac{{x + 3}}{{x - 1}}} \right) = 5$ where, $x \ne - 3,1$.
(a) $1$
(b) Cannot be determined
(c) $- \dfrac{{23}}{5}$
(d) $- 1{\text{ or }} - \dfrac{{23}}{5}$

Answer 1

The given problem revolves around the concepts of algebraic solution, which entirely depends on solving the equation mathematically such as multiplication, dividation, etc. so as to reach upto a quadratic equation i.e. $a{x^2} + bx + c = 0$ (say, firstly multiply the given equation with all the terms included in the denominator in the whole equation) and, then using the factorization method, the desired value is obtained.

Complete step-by-step solution:
Since, we have given the equation that,
$2\left( {\dfrac{{x - 1}}{{x + 3}}} \right) - 7\left( {\dfrac{{x + 3}}{{x - 1}}} \right) = 5$
As a result, to find the respective solution of the problem, multiplying the whole equation by $\left( {x + 3} \right)\left( {x - 1} \right)$ respectively, we get
$2\left( {\dfrac{{x - 1}}{{x + 3}}} \right) \times \left( {x + 3} \right)\left( {x - 1} \right) - 7\left( {\dfrac{{x + 3}}{{x - 1}}} \right)\left( {x + 3} \right)\left( {x - 1} \right) = 5\left( {x + 3} \right)\left( {x - 1} \right)$
Solving the equation mathematically, we get
$2\left( {x - 1} \right)\left( {x - 1} \right) - 7\left( {x + 3} \right)\left( {x + 3} \right) = 5\left( {x + 3} \right)\left( {x - 1} \right)$
Multiplying the above equation drastically, we get
$2\left( {{x^2} - x - x + 1} \right) - 7\left( {{x^2} + 3x + 3x + 9} \right) = 5\left( {{x^2} - x + 3x - 3} \right)$
Hence, the equation becomes
$2{x^2} - 2x - 2x + 2 - 7{x^2} - 21x - 21x - 63 = 5{x^2} - 5x + 15x - 15$
Since, the equation exists the highest power as ‘$2$’
Hence, making it fully quadratic by solving it algebraically, we get
$- 5{x^2} - 46x - 61 = 5{x^2} + 10x - 15$
$\therefore - 5{x^2} - 5{x^2} - 10x - 46x - 61 + 15 = 0$
Algebraically, contrasting the equation, we get
$- 10{x^2} - 56x - 46 = 0$
As a result, multiplying the above equation by negative ($-$) sign that is ‘$- 1$’, we get
$10{x^2} + 56x + 46 = 0$
Dividing the equation by $2$ so as to reduce the further complexity of the solution, we get
$5{x^2} + 28x + 23 = 0$
Now,
Since, we have asked to solve the given equation which seems to get the respective value of the variable in that equation particularly that is ‘$x$’,
Hence, using the factorization formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
As a result, substituting the respective values in the equation, we get
$x = \dfrac{{ - 28 \pm \sqrt {{{28}^2} - 4\left( 5 \right)\left( {23} \right)} }}{{2 \times 5}}$
Solving the equation mathematically, we get
$x = \dfrac{{ - 28 \pm \sqrt {784 - 460} }}{{10}}$
$x = \dfrac{{ - 28 \pm \sqrt {324} }}{{10}}$
Since, after factorization any quadratic equation it might exists two possible values,
Similarly, this one also deserves the two values that is
$x = \dfrac{{ - 28 + \sqrt {324} }}{{10}}$ Or, $x = \dfrac{{ - 28 - \sqrt {324} }}{{10}}$
Since, solving it mathematically i.e. taking its square root (where $324$ is the exact square root of $18$), we get
$x = \dfrac{{ - 28 + 18}}{{10}}$ Or, $x = \dfrac{{ - 28 - 18}}{{10}}$
$x = \dfrac{{ - 10}}{{10}}$ Or, $x = \dfrac{{ - 46}}{{10}}$
Hence, the required value is
$x = - 1$ Or, $x = - \dfrac{{23}}{5}$
But, we have given that the equation $2\left( {\dfrac{{x - 1}}{{x + 3}}} \right) - 7\left( {\dfrac{{x + 3}}{{x - 1}}} \right) = 5$ does not exists at $x = - 3,1$ as well
But, both the values we have solved may satisfy the given equation, as it contradicts with the given condition i.e. $x \ne - 3,1$ respectively.
$\therefore \Rightarrow$ The option (d) is absolutely correct.

Note: One must able to know the basic knowledge of algebraic terms which includes the variables x, y, z,… respectively, which seems the complexity of the respective problem particularly by knowing certain type of expansion formulae i.e. ${\left( {a + b} \right)^2}$, ${\left( {a - b} \right)^2}$, ${\left( {a + b} \right)^3}$, etc. Remember that to factorize such values as a result, it can be factorize by two methods i.e. by splitting or, extracting the middle term of the quadratic equation (if the condition exists i.e. when the factors are real and exact roots in the respective equation) otherwise directly using the factorization formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, so as to be sure of our final answer.